Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{3x - 21}{x^2 - 16x + 60} \div \dfrac{-x^2 - x}{-5x^3 + 45x^2 + 50x} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{3x - 21}{x^2 - 16x + 60} \times \dfrac{-5x^3 + 45x^2 + 50x}{-x^2 - x} $ First factor out any common factors. $a = \dfrac{3(x - 7)}{x^2 - 16x + 60} \times \dfrac{-5x(x^2 - 9x - 10)}{-x(x + 1)} $ Then factor the quadratic expressions. $a = \dfrac {3(x - 7)} {(x - 10)(x - 6)} \times \dfrac {-5x(x - 10)(x + 1)} {-x(x + 1)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {3(x - 7) \times -5x(x - 10)(x + 1) } { (x - 10)(x - 6) \times -x(x + 1)} $ $a = \dfrac {-15x(x - 10)(x + 1)(x - 7)} {-x(x - 10)(x - 6)(x + 1)} $ Notice that $(x - 10)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-15x\cancel{(x - 10)}(x + 1)(x - 7)} {-x\cancel{(x - 10)}(x - 6)(x + 1)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $a = \dfrac {-15x\cancel{(x - 10)}\cancel{(x + 1)}(x - 7)} {-x\cancel{(x - 10)}(x - 6)\cancel{(x + 1)}} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $a = \dfrac {-15x(x - 7)} {-x(x - 6)} $ $ a = \dfrac{15(x - 7)}{x - 6}; x \neq 10; x \neq -1 $